integral calculus logarithmic functions

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annnarose's picture
annnarose
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Joined: Apr 20 2015 - 6:37am
integral calculus logarithmic functions

Pano po naging sagot sa problem na integral of dx all over x(1+x^2) e 1/2 ln x^2 over 1+x^2 +c

Jhun Vert's picture
Jhun Vert
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Joined: Oct 12 2008 - 3:39pm

hindi po malinaw kung ano talaga ang equation. You can use symbols of grouping (brackets) para mas madali maintindihan.

Jeanvill Paladan Milla's picture
Jeanvill Palada...
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Joined: Jan 14 2016 - 8:23am

patulong nga po. 3x^4+4x^3+16x^2+20x+9 over (x+20)(x^2+3)^2 pasagot nga po yan hirap po kase

Jhun Vert's picture
Jhun Vert
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Joined: Oct 12 2008 - 3:39pm

Ano ang tanong dito? Mahirap manghula kuna ano ang ibig mong sabihin.

Jhun Vert's picture
Jhun Vert
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Joined: Oct 12 2008 - 3:39pm

Binasa ko uli ang post mo at naintindihan ko na. You ask why
$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \, \ln \left( \dfrac{x^2}{1 + x^2} \right) + C$

am I right? If so, here is the detail:
 

$\dfrac{1}{x(1 + x^2)} = \dfrac{A}{x} + \dfrac{Bx + C}{1 + x^2}$

$1 = A(1 + x^2) + Bx^2 + Cx$
 

When x = 0, A = 1
Equate x2: 0 = A + B, B = -1
Equate x: C = 0
 

Thus,
$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \int \left( \dfrac{1}{x} - \dfrac{x}{1 + x^2} \right) \, dx$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \int \dfrac{dd}{x} - \dfrac{1}{2} \int \dfrac{2x \, dx}{1 + x^2}$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \ln x - \dfrac{1}{2} \ln (1 + x^2) + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \left[ 2\ln x - \ln (1 + x^2) \right] + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \left[ \ln x^2 - \ln (1 + x^2) \right] + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \, \ln \left( \dfrac{x^2}{1 + x^2} \right) + C$

as we expect it to be.